Problem: The product of three consecutive, odd positive integers is seven times their sum. What is the sum of the three integers?
Explanation: Let the middle number be $n$. Then, the three consecutive odd positive integers are $n-2, n$, and $n+2$. Their sum is equal to $(n-2) + n + (n+2) = 3n$, and their product is $(n-2)n(n+2)$. Thus, $$(n-2)n(n+2) = 7 \cdot 3n = 21n.$$ Since $n \neq 0$, we can divide both sides by $n$ to obtain that $(n-2)(n+2) = 21$. By the difference of squares factorization, $21 = (n-2)(n+2) = n^2 - 4 \Longrightarrow n^2 = 25$, so $n = 5$ (as $n$ is positive). The sum of the three integers is $3n = \boxed{15}$.